Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(t^4+5t,t^3+2)$. What is the magnitude of the particle's velocity vector at $t=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5$ (Choice B) B $2$ (Choice C) C $3\sqrt{2}$ (Choice D) D $\sqrt{13}$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(t^4+5t,t^3+2)$. We are asked to find the magnitude of the particle's velocity vector at $t=0$. In other words, we need to find $||\vec{v}(0)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(t^4+5t),\dfrac{d}{dt}(t^3+2)\right) \\\\ &=(4t^3+5,3t^2) \end{aligned}$ Finding $\vec{v}(0)$ $\begin{aligned} \vec{v}({0})&=(4({0})^3+5,3({0})^2) \\\\ &=(5,0) \end{aligned}$ Finding $||\vec{v}(0)||$ $\begin{aligned} ||\vec{v}(1)||&=||(C{5},{0})|| \\\\ &=\sqrt{(C{5})^2+({0})^2} \\\\ &=\sqrt{25} \\\\ &=5 \end{aligned}$ In conclusion, the magnitude of the particle's velocity vector at $t=0$ is $5$.